Question

A solution is prepared by dissolving 10g of haemoglobin in enough water to make up
1dm^3 in volume. Calculate molarity of this solution. Molar mass of haemoglobin is 6.51x10^4g/mole.​

Answer 1

Answer:

• Molarity of solution = 1.5 × 10⁻⁴ Mol/L

Explanation:

• Mass of Haemoglobin dissolved = 10 g
• Volume of solution, V = 1 dm³ = 1 L
• Molar mass of Haemoglobin = 6.51 × 10⁴ g/mole

We need to find

• Molarity of solution, M =?

Let us first calculate the number of moles (n) of haemoglobin (solute)

→ n = mass of substance / molar mass

→ n = 10 / (6.51 × 10⁴)

→ n = 1.5 × 10⁻⁴

Using the formula for molarity of solution

→ M = n / V

[ M = Molarity, n = number of moles, V = volume of solution in litres ]

→ M = (1.5 × 10⁻⁴) / 1

→ M = 1.5 × 10⁻⁴ Mol/L

Therefore,

Molarity of solution will be 1.5 × 10⁻⁴ Mol/L.

Answer 2

$\Large{\underline{\bf{\color{cyan}GiVeN,}}}$

• Mass of dissolving haemoglobin is 10 g.

$\longmapsto\:\:\bf\orange{Mass\:(m)\:=\:10\:g}$

• Volume of water is 1 dm³.

$\longmapsto\:\:\bf\green{Volume\:(V)\:=\:1\:dm^3\:=\:1\:L}$

• Molar mass of haemoglobin is 6.51 × 10⁴ g/mol.

$\longmapsto\:\:\bf\purple{Molar\:mass\:(M)\:=\:6.51\times{10^4}\:g/mol}$

$\Large{\underline{\bf{\color{coral}To\:FiNd,}}}$

• Molarity of the solution.

$\Large{\underline{\bf{\color{lime}CaLcUlAtIoN,}}}$

$\bf\pink{We\:know\;that,}$

➣ No. of moles of any substance present in a sample is the division of given mass of the substance to it's molar mass.

$\red\bigstar\:\:{\underline{\green{\boxed{\bf{\purple{No.\:of\:moles\:(n)\:=\:\dfrac{Mass\:(m)}{Molar\:mass\:(M)}\:}}}}}}$

$:\implies\:\:\bf{n\:=\:\dfrac{10\:g}{6.51\times{10^4}\:g/mol}\:}$

$:\implies\:\:\bf{n\:=\:\dfrac{1}{6.51\times{10^3}}\:mol}$

$:\implies\:\:\bf\orange{n\:=\:0.153\times{10^{-3}}\:mol}$

$\bf\red{Again,}$

➣ Molarity is defined as the number of moles of solute present in one litre of solution.

➣ Molarity is denoted by M.

$\pink\bigstar\:\:{\underline{\purple{\boxed{\bf{\blue{M\:=\:\dfrac{No.\:of\:moles\:of\:solute\:(n)}{Volume\:of\:solution\:in\:litres\:(V)}\:}}}}}}$

$:\implies\:\:\bf{M\:=\:\dfrac{0.153\times{10^{-3}}\:mol}{1\:L}\:}$

$:\implies\:\:\bf\green{M\:=\:0.153\times{10^{-3}}\:mol/L\:}$

━─━─━─━─━─━─━─━─━─━─━─━─━─━─━

$\Large\bf\purple{Therefore,}$

➛ Molarity of the solution is 0.153 × 10⁻³ mol/L.