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:
$$
\begin{array}{cc}
\pu{1 L} &amp; -\quad \pu{0.5 mol} \\
\pu{100 mL} &amp; -\quad \pu{0.05 mol}
\end{array}
$$
$$m = \pu{0.05 mol} \times \pu{142 g/mol} = \pu{7.1 g}$$

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

$x = \pu{47.3 g}$ 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?
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.