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Calculate the mass of NaCl (molar mass = 58.5 g "mol"^(-1)] to be dissolved in 37.2 g of water to lower the freezing point by 2^@C, assuming that NaCl undergoes complete dissociation. |
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Answer» Solution :APPLY the equation, `M_2 = (K_f xx w_2 xx 1000)/(Delta T_f xx w_1)` `58.5 = (1.86 xx w_2 xx 1000)/(2 xx 37.2) ` `w_2 = (58.5 xx 2 xx 37.2)/(1.86 xx 1000) = 2.34 g ` NaCl molecule ionises to give `Na^+` and `Cl^-`ions. Thus, ONE PARTICLE gives TWO particles in solution. ` THEREFORE ` Mass of NaCl required` = 1/2 xx 2.34 = 1.17 g` |
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