How many gram equivalents of H 2 SO 4 are present in 200 ml of N -10 H 2 SO 4 solution -A- 0-01B- 0-03C- 0-04D- 0-02

The correct option is D 0.02Gram equivalents = N × volume of solution in L where N is the normality of the solution nbsp; = 110×2001000 nbsp;= nbsp;150 ...

You visited us 4 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Concentration Terms - An Introduction (w/w etc)

How many gram...

Question

How many gram equivalents of $H_{2} S O_{4}$ are present in 200 ml of $\frac{N}{10}$ $H_{2} S O_{4}$ solution ?

0.01

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.03

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.02

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.04

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 0.02
Gram equivalents = N $\times$ volume of solution in L
where N is the normality of the solution
= $\frac{1}{10} \times \frac{200}{1000}$ = $\frac{1}{50}$
= 0.02 gram equivalent of $H_{2} S O_{4}$

Suggest Corrections

Similar questions

Q. How many gram equivalents of

H_{2} S O_{4}

are present in 200 ml of

\frac{N}{10}

H_{2} S O_{4}

solution ?

Q. How many grams of

H_{2} S O_{4}

are present in

0.25 g

mole of

H_{2} S O_{4}

Q. How many grams of

H_{2} S O_{4}

are there in

1, 000

grams of a

10 %

solution? Given

(1

mol of

H_{2} S O_{4} = 98

)

Q. Find the number of milli equivalents of

H_{2} S O_{4}

present in 10 mL of N/2

H_{2} S O_{4}

solution.

Q. How many grams of sulfur (

S

) are present in

1

mole of

H_{2} S O_{4}

Join BYJU'S Learning Program

How many gram equivalents of H2SO4 are present in 200 ml of N10 H2SO4 solution ?

The correct option is C 0.02Gram equivalents = N × volume of solution in L where N is the normality of the solution = 110×2001000 = 150 = 0.02 gram equivalent of H2SO4

How many gram equivalents of $H_{2} S O_{4}$ are present in 200 ml of $\frac{N}{10}$ $H_{2} S O_{4}$ solution ?

The correct option is C 0.02
Gram equivalents = N $\times$ volume of solution in L
where N is the normality of the solution
= $\frac{1}{10} \times \frac{200}{1000}$ = $\frac{1}{50}$
= 0.02 gram equivalent of $H_{2} S O_{4}$