Question

The number of moles of solute present in the
solution of I, II and III is respectively.
1) 500mL of 0.2M NaOH
II) 200mL of 0.1N H2SO4
III) 69 of urea in 1kg of water
1 0.1, 0.01, 0.2
2 0.1, 0.01, 0.2
3 0.1, 0.01, 0.1
4 0.2, 0.01, 0.1​

Answer 1

Answer:

molality =

36

18×100

molality = 500m

Hence, the molality of H

2

SO

4

solution is 500m.

Answered By

toppr

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