Question

The amount of Na2CO3 present in 100ml of 0.1M Na2CO3 solution is ​

Answer 1

$\huge\mathbb{\blue{QUESTION-}}$

The amount of $\sf{Na_2CO_3}$ present in 100ml of 0.1M $\sf{Na_2CO_3}$ solution is

$\huge\mathbb{\blue{SOLUTION-}}$

$\large\underline{\underline{\sf Given:}}$

• Volume of $\sf{Na_2CO_3}$ (V) = 100mL = 1 L
• Molarity of $\sf{Na_2CO_3}$ (M)=0.1M

$\large\underline{\underline{\sf To\:Find:}}$

• Amount of $\sf{Na_2CO_3}$.

We know ,

$\large{\boxed{\sf Molarity (M) = \dfrac{Mole}{Volume (in\:litre)} }}$

$\implies{\sf 0.1=\dfrac{n}{0.1}}$

$\implies{\sf n=0.01 }$

$\implies{\sf n=10^{-2}}$

$\large{\boxed{\sf Mole(n)=\dfrac{Weight}{Molecular\: Weight}}}$

Molecular Weight of $\sf{Na_2CO_3}$ = 2(23)+12+3(16)

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀= 106

$\implies{\sf 10^{-2}=\dfrac{Weight}{106} }$

$\implies{\sf Weight = 10^{-2}×106 }$

$\implies{\bf \red{Weight=1.06\:g} }$

$\huge\mathbb{\blue{ANSWER-}}$

Amount of $\sf{Na_2CO_3}$ is ${\bf \red{1.06\:g}}$