209 -- A pure sample of $\mathrm{Na_2SO_4}$ has a molar mass of 88 and contains 10 and 15 percent by mass of water. Due to water absorption, the mass of the sample increases to 20 percent. The percent by mass of this salt under new conditions is how much, and if the initial sample mass is 35.5 grams, after complete reaction with barium chloride, how many grams of precipitate forms in water? (Read options from right to left; $\mathrm{BaCl_2(aq)}$ does not react with impurities; $O=16,\ Na=23,\ S=32,\ Ba=137\ \mathrm{g.mol^{-1}}$) $$\mathrm{Na_2SO_4(aq) + BaCl_2(aq) \rightarrow 2NaCl(aq) + BaSO_4(s)}$$ (1) $51.26$ , $78.2$ (2) $51.26$ , $74.9$ (3) $85.22$ , $78.2$ (4) $85.22$ , $74.9$
\textbf{209 --} A pure sample of $\mathrm{Na_2SO_4}$ has a molar mass of 88 and contains 10 and 15 percent by mass of water. Due to water absorption, the mass of the sample increases to 20 percent. The percent by mass of this salt under new conditions is how much, and if the initial sample mass is 35.5 grams, after complete reaction with barium chloride, how many grams of precipitate forms in water? (Read options from right to left; $\mathrm{BaCl_2(aq)}$ does not react with impurities; $O=16,\ Na=23,\ S=32,\ Ba=137\ \mathrm{g.mol^{-1}}$)
$$\mathrm{Na_2SO_4(aq) + BaCl_2(aq) \rightarrow 2NaCl(aq) + BaSO_4(s)}$$
(1) $51.26$ , $78.2$ \quad (2) $51.26$ , $74.9$
(3) $85.22$ , $78.2$ \quad (4) $85.22$ , $74.9$
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