Question

The weight of solute present in 200ml of 0.1M H2SO4 is

Answer 1

hey!! Here is your answer
Given,
V= 200ml
M=0.1
Molecular weight of H2SO4= 2(1)+32+4(16)
=98
Formula :
Molarity, M = W/Mol.wt *1000/V in ml
0.1 = W/98*1000/200
W= 98*0.1*1/5
W= 49/25
W=1.96
Weight of the solute present = 1.96g

Answer 2

Answer : The weight of solute present in 200 ml of 0.1 M $H_2SO_4$ is, 1.96 g

Solution : Given,

Volume of solution = 200 ml

Molarity of $H_2SO_4$ solution = 0.1 M = 0.1 mole/L

Molar mass of $H_2SO_4$ = 98 g/mole

Molarity : It is defined as the number of moles of solute present in one liter of solution.

Formula used :

$Molarity=\frac{w\times 1000}{M\times V_s}$

where,

w = mass of $H_2SO_4$ (solute)

M = molar mass of $H_2SO_4$

$V_s$ = volume of solution in liter

Now put all the given values in the above formula, we get the mass of solute.

$0.1mole/L=\frac{w\times 1000}{(98g/mole)\times (200L)}$

$w=1.96g$

Therefore, the weight of solute present in 200 ml of 0.1 M $H_2SO_4$ is, 1.96 g