Calculate the molality of a solution containing 14.3 g of nacl in 42.2 g of water.
a. 2.45 ï´ 10â4 m
b. 5.80 ï´ 10â4 m
c. 2.45 ï´ 10â1 m
d. 103 m
e. 5.80 m

Respuesta :

molarity =Number of moles of compound*1000/(mass of solvent)

Number of moles of NaCl=Given mass/molar mass

Given mass of NaCl=14.3 g

molar mass of NaCl=58.44g

Number of moles of NaCl= 14.3/58.44

=0.244

Mass of solvent (Water)=42.2 g

Molarity=0.244*1000/(42.2)

= 5.80m

Answer:  e) 5.80 m

Explanation:

Molality : It is defined as the number of moles of solute present per kg of solvent

Formula used :

[tex]Molarity=\frac{n\times 1000}{W_s}[/tex]

where,

n= moles of solute

To calculate the moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{14.3g}{58.5}=0.24moles[/tex]

[tex]W_s[/tex] = weight of solvent in g= 42.2 g

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

[tex]Molarity=\frac{0.24moles\times 1000}{42.2g}=5.80mole/kg[/tex]

Therefore, the molality of solution will be 5.80 m