A) \[1.043\text{ }m\]
B) \[0.143m\]
C) 10.43 m
D) none of these
Correct Answer: C
Solution :
93% \[{{H}_{2}}S{{O}_{4}}\] (w/V) means 93 g of \[{{H}_{2}}S{{O}_{4}}\] is present in 100 cc of solution. Since volume of solution =1000 mL \[\therefore \].Amount of \[{{H}_{2}}S{{O}_{4}}\] = 930 g per 1000 mL Density =1.84 g/mL \[D=\frac{M}{V}\] \[M=D\times V\] \[=1.84\times 1000\] \[=1840g\] weight of solvent \[=1840-930\] \[=910g\] \[\therefore \] molality \[=\frac{930}{98}\times \frac{1000}{910}\] \[=\frac{930000}{89180}\] \[=10.43m\]
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