Question

I need help ASAP plsss!!! ​

Answer 1

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

Step-by-step 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=(Molarity* volume)/(1000)`

`M_(NaOH)= 0.10`

`V_(NaOH) = 20.0cm^(3)`

`moles=(0.10* 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* 20.0=M_(HCl)* 12.5`

`M_(HCl)=(0.10* 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`