Question

calculate molarity of the solution containing 0.04 g of HCl in 250ml of water?​

Answer 1

Given :

▪ Weight of HCl = 0.04g

▪ Volume of water = 250ml

To Find :

▪ Molarity of the solution.

Concentration of solution :

❇ Name : Molarity

❇ Symbol : M

❇ Definition : Number of moles of solute dissolved one litre of solution.

❇ Effect of trmperature : No effect

Mathematically,

$\bigstar\:\underline{\boxed{\bf{\red{M=\dfrac{Moles\:of\:solute}{Volume\:of\:solution\:in\:L}}}}}$

▪ Molecular weight of HCl = 36.5g/mole

Conversion :

↗ 1000ml = 1L

↗ 250ml = 250/1000 = 0.25L

Calculation :

$\dashrightarrow\sf\:M=\dfrac{Moles\:of\:solute}{Volume\:of\:solution\:in\:L}\\ \\ \dashrightarrow\sf\:M=\dfrac{W_{solute}}{{(M_w)}_{solute}\times V_{solution}}\\ \\ \dashrightarrow\sf\:M=\dfrac{0.04}{36.5\times 0.25}\\ \\ \dashrightarrow\underline{\boxed{\bf{\blue{M=0.004\:gL^{-1}}}}}\:\orange{\bigstar}$

Answer 2

Given:-

❐ weight of the HCl = 0.04 g.

❐ Volume of water = 250 ml = 250/1000 = 0.25 lit

To Find:-

Molarlity of the solution

Concentration of solution:-

Name :- Molarlity

symbol:- m

Temperature:- No effect.

$\large\underline{\underline{\boxed{\mathtt{\color{teal}{M = \frac{moles \: of \: solute}{Volume \:of\: in \: L}}}}}}{\color{olive}{\bigstar}}$

Calculation:-

$\large\underline{\underline{\boxed{\mathtt{\color{red}{M = \frac{moles \: of \: solute}{Volume \:of\: in \: L}}}}}}{\color{olive}{\bigstar}}$

$\large{\implies{\tt{\frac{0.04}{36.5×0.25}}}}$

$\longrightarrow\boxed{\tt{0.004gL^-1}}$