Question

Calculate the mass of the solute in the following
solutions :
(i)100 cm^3 of N/10 KOH
(ii) 250 cm^3 of NH2SO4
(iii) 250 cm^3 of 2M HNO3
(iv) 150 cm^3 of M/2 HCl
​

Answer 1

Answer:

1)

N= 1/10

N = M * x ( where is x valancy factor)

1/10 = M * 1   ( as only 1 OH ion is present )

so M= 1/10

M =  no. of moles / vol. of the solution

1/10 = given mass / molecular mass * vol. of solution

1/10 = given mass*1000 / 56*100    (as 1cm³= 1 / 1000 Litres )

given mass = 56 / 10*10 = 0.56 g

2)

N = 1

N = M * x ( where is x valancy factor)

1 = M * 2 ( as 2 H ions are present in h2so4)

M = 1/2

M =  no. of moles / vol. of the solution

1/2 =  given mass / molecular mass * vol. of solution

1/2 = given mass*1000/ 98 * 250      ( mol. mass of h2so4 = 98)

given mass =  98 / 2*4 = 12.25 g

3)

M = 2

M =  no. of moles / vol. of the solution

M =  given mass / molecular mass * vol. of solution

2 =  given mass*1000 / 63*250      ( mol. mass of hno3 is 63)

given mass = 63/2 = 31.5 g

4)

M = 1/2

M =  no. of moles / vol. of the solution

M =  given mass / molecular mass * vol. of solution

1/2 = given mass*1000 / 36.5*150

given mass = 15*36.5/100*2 = 7.5*36.5/100 = 2.73 g

in the above questions M is molarity N is normality

for making your understanding clear you can  refer physics wallah chemistry chapter  1 class 11 ( lecture 4 and 5 ) (old)

hope it help

Answer 2

Answer:

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