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Questions  

3.15 grams of solute is present in 200ml of 0.25 M solution. The solute may be

a
H2C2O4
b
HCl
c
H2SO4
d
HNO3

detailed solution

Correct option is D

Molarity, M = no. of moles of solutevol. of solution (in L)

no. of moles of solute, n = given massmolar mass

Molarity of a solution is defined as the amount of solute dissolved in 1L of solution.

M = 0.25 M, grams of solute = 3.15 g ,vol. of solution = 200 mL = 0.2 L

So,  0.25 = 3.15MM0.2 ; molecular mass = 63 g/mol

H2C2O4 = 90 g/mol ;HCl = 36.5 g/mol ; H2SO4 = 98 g/mol ; HNO3 = 63 g/mol

Hence, correct option is: (D) HNO3

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