Question

calculate the amount of oxalic acid required to prepare 1000 mL 0.1 M standard solution of oxalic acid​

Answer 1

Answer: 9 grams

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

Molarity = 0.1 M

n= moles of solute  

$V_s$ = volume of solution in ml = 1000 ml

Now put all the given values in the formula of molarity, we get

$0.1=\frac{n\times 1000}{1000}$

n = 0.1

$moles of solute=\frac{\text {given mass}}{\text {molar mass}}=\frac{xg}{90g/mol}$

$0.1=\frac{xg}{90g/mol}$

$x=9g$

Thus amount of oxalic acid required to prepare 1000 mL 0.1 M standard solution of oxalic acid​ is 9 grams

Answer 2

Answer: 9 grams

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

Molarity=\frac{n\times 1000}{V_s}Molarity=

V

s

n×1000

where,

Molarity = 0.1 M

n= moles of solute

V_sV

s

= volume of solution in ml = 1000 ml

Now put all the given values in the formula of molarity, we get

0.1=\frac{n\times 1000}{1000}0.1=

1000

n×1000

n = 0.1

moles of solute=\frac{\text {given mass}}{\text {molar mass}}=\frac{xg}{90g/mol}molesofsolute=

molar mass

given mass

=

90g/mol

xg

0.1=\frac{xg}{90g/mol}0.1=

90g/mol

xg

x=9gx=9g