Question

Equal moles of h2o and nacl are present in a solution hence molality of nacl solution is

Answer 1

Consider ‘Y’ mol of water and ‘Y’ mol of NaCl are mixed to prepare a solution.

1 mol of water contains 18 g

That means, Y mol of water contains 18Y g

Mass / Density = Volume

Density of water = 1

So, volume of Y mol water = 18Y / 1 = 18Y mL

That means Y mol of NaCl is dissolved in 18Y mL of water.

1 mol of NaCl contains 58.44 g

That means, Y mol of NaCl contains 58.44Y g

Molality can be calculated as follows:

Molality = (weight of NaCl x 1000) / (Equivalent weight of NaCl x volume of water)

Substitute in equation,

Molality = (58.44Y x 1000) / (58.44 x 18Y)

Molality = 55.56 M

Molality of NaCl solution is 55.56 M.

Answer 2

Thank you for this question. Please find the answer below:

Equal moles of h2o and nacl are present in a solution hence molality of NaCl solution is 55.55 mol/ litre .

Explanation:

Molarity = moles of solute / volume of solution

If we consider the solution as not saturated then we can ignore any changes in final volume after adding NaCl , otherwise it decreases in case of saturated solution .

At NTP density of water is 1000 kg/kl or

1000 gram/litre

and molecular mass of H2O = 18 grams

Density = mass / volume

Mole = mass / molar mass

Volume of water will be = mass of H2O/ density

Mass of H2O = Moles of H2O * Molar mass of H2O

= ( X *18 )

Volume = 18 X / 1000 litre

Thus molarity will be

= X / { ( 18* X )/ 1000 } mole / litre

= 1000/ 18

= 55.55 mol/ litre .