# Question ee998

Apr 26, 2015

You can use the molarity of the solution to determine how many moles of dipotassium phosphate you'd get in 500 mL, then multiply that value by the molar mass of the compound

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{{K}_{2} H P {O}_{4}} = 0.4 \text{moles"/cancel("L") * 500 * 10^(-3)cancel("L") = "0.2 moles }$ ${K}_{2} H P {O}_{4}$

and

0.2cancel("moles "K_2HPO_4) * "174.17 g"/(1cancel("mole "K_2HPO_4)) = "34.83 g "# ${K}_{2} H P {O}_{4}$

If you take into account sig figs, the answer should be

${m}_{{K}_{2} S {O}_{4}} = \text{30 g}$ $\to$ since you only give one sig fig for the volume and molarity of the dipotassium phosphate.