Question

The molality of 1L solution of 93% H2SO4(w/v) having density 1.84 g/mL is


(A)1.043 m
(B)0.143 m
(C)10.43 m
(D)none of these

Answer 1

Answer:

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Answer 2

Answer:C

Explanation:

Since volume is 1000ml, amount of H2SO4 is 930 g/litre. Also density is given 1.84g/ml M= density x volume = 1.84 x 1000 = 1840 g

wt of solvent = 1840 - 930 = 910 g

= So, molality = 930/98 x 1000/910 = 10.43