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The amount of oxalic acid (mol. wt. 63) ...

The amount of oxalic acid (mol. wt. 63) required to prepare 500 mL of its 0.10 N solution is

A

0.315 g

B

3.150 g

C

6.300 g

D

63.00 g

Text Solution

Verified by Experts

The correct Answer is:
B
1000 mL of 1 N oxalic acid sol = 63 g
500 mL of 0.1 N oxalic acid sol `=63/1000 xx 500 xx 0.1 = 3.15 g`
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Knowledge Check

  • The amount of oxalic acid (eq. wt 63) required to preare 500 ml of its 0.10 N solution is

    A
    `0.315g`
    B
    `3.150g`
    C
    `6.300g`
    D
    `63.00g`

  • The amount of oxalic acid required to prepare 300 mL of 2.5 M solution is : (molar mass of oxalic acid = 90 g mol^(–1) ) :

    A
    67.5 g
    B
    9.45 g
    C
    6.75 g
    D
    94.5 g

  • The mole fraction of oxalic acid (Molar mass 63) required to prepare 0.10 m solution in water is

    A
    1
    B
    0.0018
    C
    6.3
    D
    0.0992

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    Amount of glucose ( C_(6)H_(22)O_(6) ) required to prepare 100 mL of 0.1 M solution is:

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