Chapter 3 - Chemical Equations and Reaction of Stoichiometry - Exercises - Concentrations of Solutions - Percent by Mass - Page 110: 57

a) $n(K_2Cr_2O_7) = 0.2551mol$ b) $m(K_2Cr_2O_7) = 75g$ c) $m(H_2O)= 425g$

a) $n(K_2Cr_2O_7) = \frac{m(K_2Cr_2O_7)}{M(K_2Cr_2O_7)} = \frac{\omega \cdot m(solution)}{M(K_2Cr_2O_7)} = \frac{0.15\cdot 500g}{294\frac{g}{mol}}=0.2551mol$ b) $m(K_2Cr_2O_7) = \omega \cdot m(solution) = 0.15\cdot 500g = 75g$ c) $m(H_2O) = m(solution) - m(K_2Cr_2O_7) = 500g - 75g = 425g$