Chemistry: What mass of kbr is present in 25 ml of a 0.85 m solution of potassium chloride?

What mass of kbr is present in 25 ml of a 0.85 m solution

toshahoskins0098

13.03.2022

Answer : The correct answer for mass of KBr = 2.53 g

Given :

Molarity of KBr solution = 0.85 M

Volume of KBr solution = 25 mL

Converting volume from mL to L ( 1 L = 1000 mL )

$Volume of solution = 25 mL * \frac{1 L }{1000mL}$

Volume of solution = 0.025 L

Mass of KBr = ?

Mass of KBr can be calculated using following steps :

1) To find mole of Kbr :

Mole of KBr can be calculated using molarity .

Molarity : It is defined as mole of solute present in volume of solution in Liter .

It uses unit as M or $\frac{mol}{L}$

It can be expressed as :

$Molarity = \frac{mol of solute (mol)}{volume of solution (L)}$

Plugging value of molarity and volume

$0.85 \frac{mol}{L} = \frac{mol of Kbr}{0.025 L}$

Multiplying both side by 0.025 L

$0.85 \frac{mol}{L} * 0.025 L = \frac{mole of KBr}{0.025 L} * 0.025 L$

Mole of KBr = 0.02125

2) To find mass of Kbr :

Mass of Kbr can be calculated using mole . Mole can be expressed as :

$Mole (mol) = \frac{mass (g) }{molar mass \frac{g}{mol} }$

Mole of Kbr = 0.02125 mol

Molar mass of KBr = 119.00 $\frac{g}{mol}$

Plugging values in mole formula

$0.02125 mol = \frac{mass (g)}{119.00 \frac{g}{mol}}$

Multiplying both side by 119.00 $\frac{g}{mol}$

$0.02125 mol * 119.00 \frac{g}{mol} = \frac{mass (g)}{119.00 \frac{g}{mol}} * 119.00\frac{g}{mol}$

Mass of KBr = 2.53 g

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