Question

250 mL of 0.5 M NaOH was added to 500 mL of 1 M HCl. The number of unreacted HCl molecules in the solution after complete reaction is ______ × 10²¹. (Nearest integer) (N_A = 6.022 × 10²³).

पर्याय

• 108

• 66

• 85

• 266

Answer 1

250 mL of 0.5 M NaOH was added to 500 mL of 1 M HCl. The number of unreacted HCl molecules in the solution after complete reaction is 266 × 10²¹.

Explanation:

Given,

V_NaOH = 250 ml

[NaOH] = 0.5 M

V_HCl = 500 ml

[HCl] = 1 M

N_A = 6.022 × 10²³

The molarity and the number of moles of substance are related to each other by given expression.

M = `"n"/"V"`

Here, M represents the molarity of the solution. The n represents number of moles and V is the volume in litres.

The number of moles can be calculated as follows:

n = M × V     ...(i)

The number of millimoles can be calculated using the same formula, only volume needs to be considered in ml. 

The number of millimoles can be expressed in a similar manner.

Number of millimoles = M × V ..... (ii) 

Here, volume is in mililitres (ml). 

Substitute the appropriate values in equation 2 to determine the millimoles of NaOH.

Number of millimoles of NaOH 

= V_NaOH × Molarity of NaOH 

Number of millimoles of NaOH = 250 × 0.5 = 125 

Similarly, millimoles of HCl can be determined. 

Number of millimoles of HCl = V_HCl × Molarity of HCI 

Number of millimoles of HCl = 500 × 1 = 500

Let us write the complete reaction between NaOH and HCL. It is a neutralisation reaction during which a salt and water is produced.

	NaOH	+	HCl	→	NaCl	+	H2O
At t = 0	125		500		0		0
At t = t	0		375		125		125

500 moles of HCl and 125 moles of NaOH were present at first. As there were no products created at the beginning of the reaction, there were exactly zero moles of products.

After sometime, 125 moles of NaOH reacts with 125 moles of HCl, producing 125 moles each of salt and water. The remaining moles of HCl are 375 moles.

Therefore, millimoles of HCl left after the reaction are 375 millimoles. 

The millimoles can be converted to moles as follows:

Moles of HCl = 375 × 10⁻³ mol

The number of molecules of HCl can be computed by multiplying Avogadro's number with the number of millimoles of HCl.

No. of HCl molecules = 6.022 × 10²³ × 375 × 10⁻³

= 225.8 × 10²¹

≈ 226 × 10²¹

Therefore, 250 ml of 0.5 M NaOH was added to 500 mL of 1 M HCl. The number of unreacted HCl molecules in the solution after complete reaction is 26 × 10²¹.