Math, asked by rohit999991, 11 months ago

DPP-3
Common salt obtained from sea water contains 8.775% NaCl by mass. The number of formu
units of NaCl present in 25 g of this salt is:
Oo22

Answers

Answered by Aman2630

Answer : 2.2586×10^22

Step-by-step explanation:

Formula mass of NaCl = 58.5

8.775 % NaCl by mass means in 100 g of common salt obtained, there is 8.775 g NaCl

Therefore in 25 g of this salt, amount of NaCl will be = (8.775/100) × (25)

= 2.19375 g

No. of formula units of NaCl = {(given mass of NaCl)÷(formula mass of NaCl)}×6.023 × 10^23

Therefore answer is

= {2.19375/58.5}x6.023×10^23

= 0.0375×6.023×10^23

=0.22586×10^23

=2.2586×10^22

Aman2630: so please

rohit999991: ok

rohit999991: i don't knows answer but sending you options

rohit999991: (a)3.367×10^23 formula units (b)2.258×10^22 (C)3.176×10^23 (D)4.73×10^25

Aman2630: I got the answer

Aman2630: its (b)

Aman2630: I forgot to multiply the no. of moles of NaCl calculated wih AVOGADRO'S NUMBER

Aman2630: Sorry, for my silly mistake. It won't happen next time.

rohit999991: it's ok . insann se bhi galti to hoti hai . not a serious problem . but thanks for reply .

Aman2630: You're welcome.

Answered by nilkanthgohil1010

Answer:

2.25 * 10^22 units

Step-by-step explanation:

Let 100g of sea water contains 8.775g of sodium chloride

So 25g of water would have 2.19375g of sodium chloride

Molar mass of sodium chloride is 35.5+23=58.5g

No of moles in 2.19375 g of sodium chloride is 2.19375/58.5 moles

No. Of molecules present would be (2.19375/58.5)*Avogadro number

= 2.19375* 6* 10^23/(58.5)

= 2.25 * 10^22 units

HOPE ITS HELP .............

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