# How many grams of sodium sulfate and water in 100ml of 0.5M solution

Calculate how many grams of sodium sulfate ($\ce{Na2SO4}$) and water ($\ce{H2O}$) should be used to prepare $\pu{370 g}$ of 15% solution of this salt. How many mL of this 15% solution should be taken to prepare $\pu{100 mL}$ of $\pu{0.5 M}$ solution of $\ce{Na2SO4}$?

My calculations:

$\pu{370 g}$ = 85% of water, 15% of sodium. So we need $\pu{370 g} \times 0.85 = \pu{314.5 g}$ of $\ce{H2O}$ and $\pu{370 g} \times 0.15 = \pu{55.5 g}$ of sodium sulfate.

We need $\pu{100 mL}$ of $\pu{0.5 M}$ solution.

The amount of sodium sulfate we need: $$\begin{array}{cc} \pu{1 L} & -\quad \pu{0.5 mol} \\ \pu{100 mL} & -\quad \pu{0.05 mol} \end{array}$$ $$m = \pu{0.05 mol} \times \pu{142 g/mol} = \pu{7.1 g}$$

To get $\pu{7.1 g}$ of sodium sulfate we need $$\begin{array}{ll} \pu{100g} & -\quad \pu{15 g} \\ x & -\quad \pu{7.1 g} \end{array}$$

$x = \pu{47.3 g}$ of solution that's over $\pu{50 mL}$ of it.

But the answer states we need $\pu{3.75 mL}$ of solution.

Could you tell me where my reasoning is wrong or it is a problem with the answer?

• I agree with your mass calculations, and the mass of sodium sulfate required as well. You should need the 47.3g you've calculated as well. To find a volume, you'd either need to know the density of the solution or make some (not insignificant) assumptions about the solution's density. In any case, the answer you were given is incorrect. – Jason Patterson Oct 24 '15 at 18:53
• The density of a 15 % aqueous solution of sodium sulfate at 20 °C is about 1.14 g/ml. – Loong Oct 27 '15 at 18:58

As Jason indicated your calculations are corrected.

• 370 grams of 15% solution of sodium sulfate does have 314.5g $\ce{H2O}$ and 55.5g $\ce{Na2SO4}$.

• Your calculation for grams of solution are correct. 47.3g of solution is required.

However the density would be expected to be somewhat above 1.00 g/ml not less. So a little less than 47.3 ml of solution would be required. However the density of the solution is nowhere near 10 g/ml so the book answer is horribly wrong.

To solve for ml of solution, I'd have just stated an assumption that the density of the solution was 1.00 g/ml since you weren't given that information. There is absolutely no way to calculate the density given the information in the problem.