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What mass of KBr is present in 75.0 mL of 0.250 M KBr solution?

Respuesta :

Answer:

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}Volumeofsolution=25mL∗

1000mL

1L

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}

L

mol

It can be expressed as :

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

volumeofsolution(L)

molofsolute(mol)

Plugging value of molarity and volume

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

L

mol

=

0.025L

molofKbr

Multiplying both side by 0.025 L

0.85 \frac{mol}{L} * 0.025 L = \frac{mole of KBr}{0.025 L} * 0.025 L0.85

L

mol

∗0.025L=

0.025L

moleofKBr

∗0.025L

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(mol)=

molarmass

mol

g

mass(g)

Mole of Kbr = 0.02125 mol

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

mol

g

Plugging values in mole formula

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

119.00

mol

g

mass(g)

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

mol

g

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

mol

g

=

119.00

mol

g

mass(g)

∗119.00

mol

g

Mass of KBr = 2.53 g