Question

I need help ASAP plsss!!! ​

Answer 1

Answer:

a). Moles of NaOH used = 0.002 moles

b). Moles of HCl used = 0.002 moles(because it is neutralisation reaction , so moles of HCl - NaOH)

c) Concentration of HCl =0. 16 M

Explanation:

Molarity(M) : It is the number of moles of the substance present in 1 liter of the solution. Molarity is expressed in the unit of moles/liter.

$1liter = 1dm^{3}$

In the given question:

0.10 mol dm-3 of Sodium Hydroxide is given.

This is Molarity of NaOH = 0.10 mole/L(because dm3 and L are same thing)

Moles of NaOH in 1 liter solution = 0.1 mole

$M_{NaOH}= 0.10$

$V_{NaOH} = 20.0cm^{3}$

Now,

1 liter = 1000 ml

$1liter = 1dm^{3}$

Put 1000 mL in place of 1 liter

$1000mL = 1dm^{3}$

1 dm = 10 cm

$1dm^{3}$= 1000cm^{3}[/tex]

replace 1 dm3 by cm3

$1000mL = 1000cm^{3}$

1 mL = 1 cm3 ,

So , 20 cm3 = 20 mL

$moles=\frac{Molarity\times volume}{1000}$

$M_{NaOH}= 0.10$

$V_{NaOH} = 20.0cm^{3}$

$moles=\frac{0.10\times 20}{1000}$

moles of NaOH used solution = 0.002 moles

$V_{HCl} = 12.5 mL$

use equation :

$M_{NaOH}V_{NaOH}=M_{HCl}V_{HCl}$

$0.10\times 20.0=M_{HCl}\times 12.5$

$M_{HCl}=\frac{0.10\times 20}{12.5}$

$M_{HCl}=0.16M$

So, 0.16 M of HCl neutralises NaOH

Concentration of HCl =0.16 M

$HCL+NaOH\rightarrow H_{2}O+NaCl$