25 g of a solute of molar mass `250 g mol^(-1)` is dissolved in 100 ml of water to obtain a solution whose density is `1.25 g ml^(-1)`. The molarity and molality of the solution are respectively

Options

A. 0.75 and 1
B. 0.8 and 1
C. 1 and 0.8
D. 1 and 0.75

Answer

C

Explanation

No. Of moles of solute = `(25)/(250)=0.1` moles
Molarlity = `("No. Of moles of solute")/("Volume of solvent in litre")=(0.1)/((100)/(1000))=1M`
Molality = `("no. Of moles of solute")/("mass of solvent in kg")`
Mass `=d xx V`
`=1.25xx100=125g`
Molality = `(0.1)/(125//1000)=0.8m`.