00:01
So we have to prepare to prepare 2 liter 0 .169 molar naoh, okay, naoh from 53 .4 weight percentage of naoh.
00:36
Okay, so we can say that the mass of neoh required therefore.
00:44
The mass of n .a .o .h required is equal to 2 multiplied by 0 .169 multiplied by 40 gram per 1 mole, which is equal to, we can say, 13 .52 gram.
01:14
Okay.
01:14
This is nah.
01:18
Okay.
01:19
So this is the mass required for noh.
01:22
And the volume of solution, therefore, volume of solution required, volume of solution required is equals to, okay, this is different, volume of solution required is equals to, we can say that, 100 divided by 53 .4, multiplied by 13 .52, 13 .52, multiplied by 1 ,3 .52, multiplied by 1 ,000, 1 ml divided by 1 .52 gram...