Question

(ntse exam question) A sample of MgCO3 contains 3.01 × 1023 Mg2+ ions and 3.01 × 1023 2 CO 3  ions. The mass of the sample is (1) 42 mg (2) 84 g (3) 0.042 kg (4) 42 mol

Answer 1

molar mass of MgCO3 = 24+12+16*3 = 84g

If it contains 3.01 × 1023 Mg2+ ions and 3.01 × 1023 2 CO 3 ions, number of moles of MgCO3 present = 0.5 molemass = 84*.05 = 42gram = 0.042kg
Answer is (3)

Answer 2

Answer: The correct answer is Option 3.

Explanation:

Magnesium carbonate is formed by the combination of $Mg^{2+}\text{ and }CO_3^{2-}$ ions

We are given:

Number of $Mg^{2+}$ ions = $3.01\times 10^{23}$

Number of $CO_3^{2-}$ ions = $3.01\times 10^{23}$

According to mole concept:

$6.022\times 10^{23}$ number of molecules occupy 1 mole of a gas.

So, $3.01\times 10^{23}$ number of magnesium ions will occupy = $\frac{1}{6.022\times 10^{23}}\times 3.01\times 10^{23}=0.5mol$ of magnesium carbonate.

To calculate the mass for the given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of magnesium carbonate = 84 g/mol

Moles of magnesium carbonate = 0.5 moles

Putting values in above equation, we get:

$0.5mol=\frac{\text{Mass of magnesium carbonate}}{84g/mol}\\\\\text{Mass of magnesium carbonate}=(0.5mol\times 84g/mol)=42g=0.042kg$

Conversion factor used:  1 kg = 1000 g

Hence, the correct answer is Option 3.