Recall how to find the molality of a solution:

\(\displaystyle \text{Molality}=\frac{\text{moles of solute}}{\text{kg of solvent}}\)

First, start by finding the moles of glucose that we have. The molar mass of glucose is \(\displaystyle 180.16\text{g/mol}\).

\(\displaystyle 10.2\text{g of glucose}\cdot\frac{\text{1 mole glucose}}{180.16\text{g of glucose}}=0.0566\text{ moles of glucose}\)

Next, convert the grams of water into kilograms.

\(\displaystyle 405\text{g}\cdot\frac{1\text{kg}}{1000\text{g}}=0.405\text{kg}\)

Now, plug in the moles of glucose and kilograms of water into the equation for molality.

\(\displaystyle \text{Molality}=\frac{0.0566}{0.405}=0.140m\)

Start by assuming that we have \(\displaystyle 1.00\text{L}\) of this solution. Recall that hydrogen peroxide has a molecular formula of \(\displaystyle \text{H}_2\text{O}_2\).

Use the given density to find the mass of the solution.

\(\displaystyle 1.00\text{L}\cdot\frac{1000\text{mL}}{1\text{L}}\cdot\frac{1.01\text{g}}{\text{mL}}=1010\text{g}\)

Next, find the mass of the hydrogen peroxide present in the solution.

\(\displaystyle 1010\text{g}(0.152)=153.52\text{g}\)

Convert the mass of hydrogen peroxide into moles of hydrogen peroxide.

\(\displaystyle 153.52\text{g of H}_2\text{O}_2\cdot\frac{1\text{ mole H}_2\text{O}_2}{31.04\text{g H}_2\text{O}_2}=4.95\text{ moles of H}_2\text{O}_2\)

Recall how to find the molarity of a solution:

\(\displaystyle \text{Molarity}=\frac{\text{moles of solute}}{\text{liters of solution}}\)

Since we have \(\displaystyle 1.00\text{L}\) of solution, the molarity is \(\displaystyle 4.95M\).

\(\displaystyle 211 g\) sodium hydrogen carbonate dissolved in a \(\displaystyle 10.0 L\) solution has \(\displaystyle 0.251 M\). 

The first step is to calculate how many moles of \(\displaystyle NaHCO_{3}\) are present.  \(\displaystyle 211 g NaHCO_{3}\frac{1 mol NaHCO_{3}}{84.006 g NaHCO_{3}}=2.51 mol NaHCO_{3}\)

We calculate molarity with the following equation: 

\(\displaystyle M = \frac{moles of solute}{liters of solution }\)

\(\displaystyle M = \frac{2.51 moles NaHCO_{3}}{10 L} = 0.251 M\)

\(\displaystyle 35 mL\) of \(\displaystyle 5.0 M CuSO_{4}\) solution are needed to prepare \(\displaystyle 0.350 L\) of \(\displaystyle 0.500 M CuSO_{4}\).  

We can use the formula \(\displaystyle M_{1}V_{1}= M_{2}V_{2}\)

Therefore, \(\displaystyle (5.00 M_{1})(V_{1})=(0.350 M_{2})(0.500 L)\)

\(\displaystyle V_{1}= 0.035 L = 35 mL\)

The relative concentrations of aluminum sulfate \(\displaystyle (Al_{2}(SO_{4})_{3})\) are

\(\displaystyle 6.0 M Al^{3+}\)

\(\displaystyle 9.0 M SO_{4}^{2-}\)

\(\displaystyle \frac{3.0 mol Al_{2}(SO_{4})_{3}}{1.0 L}\cdot\frac{2 mol Al^{3+}}{1 mol Al_{2}(SO_{4})_{3}}=6.0 M Al^{3+}\)

\(\displaystyle \frac{3.0 mol Al_{2}(SO_{4})_{3}}{1.0 L}\cdot\frac{3 mol SO_{4}^{2-}}{1 mol Al_{2}(SO_{4})_{3}}=9.0 M Al^{3+}\)

Molarity, molality, and normality are all units of concentration in chemistry. Molarity (\(\displaystyle M\)) is defined as the number of moles of solute per liter of solution. Molality (\(\displaystyle m\)) is defined as the number of moles of solute per kilogram of solvent. Normality (\(\displaystyle N\)) is defined as the number of equivalents per liter of solution. Molality, as compared to molarity, is also more convenient to use in experiments with significant temperature changes. This is because the volume of a solution increases with temperature, and heating causes molarity to decrease; however, since molality is based on masses rather than volumes, molality remains unchanged.

Molality (\(\displaystyle m\)) is defined as moles of solute per kg of solvent.

\(\displaystyle NaCl\) is the solute (it is what is being dissolved) and water is the solvent (what is doing the dissolving).

 Let’s start converting \(\displaystyle 9.5g\) of \(\displaystyle NaCl\) to moles by dividing by its molecular weight using scientific notion through the entire molality calculations as necessary to simplify calculations.

\(\displaystyle 9.5g\ NaCl*1mol\ NaCl*\frac{1}{5.84*10g\ NaCl}=0.163mol\ NaCl\)

Now we must convert the \(\displaystyle 300g\) of water to \(\displaystyle kg\) then divide \(\displaystyle mols\) \(\displaystyle NaCl\) by it to get the molality:

\(\displaystyle 3*10^{2}g\ H_2O*1kg*\frac{1}{10^3g}=3*10^{-1}kg\ H_2O\)

Molality of solution: 

\(\displaystyle =1.63*10^{-1}mol\ NaCl*\frac{1}{3*10^{-1}kg\ H_2O}\)

=\(\displaystyle .54m\)

Molarity \(\displaystyle =\) moles of solute per liter of solution.

We are given that there are \(\displaystyle 35g\) of \(\displaystyle MgCO_3$\) and \(\displaystyle 750ml\) of solution.

First lets convert the \(\displaystyle ml\) of solution to liters. It is easiest to use scientific notion in all calculations for easy simplification:

\(\displaystyle 750ml\)=\(\displaystyle 7.5*10^2ml\)

\(\displaystyle \frac{7.5*10^{2}ml*1L}{(10^3ml)}=7.5*10^{-1}L\)

Now we must convert \(\displaystyle 35g\) of \(\displaystyle MgCO_3$\) to moles to get moles of solute per liter of solution.

This is done by dividing \(\displaystyle 35g\) by its molecular weight which is \(\displaystyle 84.13g\). Once again we can use scientific notion to simplify calculations:

\(\displaystyle 3.5*10g\ MgCO_3*1mol\ MgCO_3*\frac{1}{(8.413*10g\ MgCO_3)}\)

\(\displaystyle =.416mol\) \(\displaystyle MgCO_3$\) \(\displaystyle solute\)

Now we can divide \(\displaystyle .416mol\) \(\displaystyle MgCO_3\) by \(\displaystyle 7.5*10^{-1}L\) solution to get the molarity.

\(\displaystyle 4.16*10^{-1}mol\ MgCO_3*\frac{1}{7.5*10^{-1}L}\)

\(\displaystyle =\)\(\displaystyle .55M\)

In order to solve for the volume of water needed we must use this equation:

\(\displaystyle Minitial*Vinitial=Mfinal*Vfinal\)

\(\displaystyle Minitial=7M\)

\(\displaystyle Vinitial=65ml\)

\(\displaystyle Mfinal =2M\)

\(\displaystyle Vfinal=?\)

solve for \(\displaystyle Vfinal\) since we are looking for the final volume of water needed to dilute the existing solution:

\(\displaystyle Vfinal=Minitial*Vinitial/Mfinal\)

\(\displaystyle =7M*65ml/2M\)\(\displaystyle =227.5ml\)