Question

The cost of table salt (NaCl) and table sugar ($C_{12}H_{22}O_{11}$) are Rs. 2 per kg and Rs. 6 per kg respectively calculate their cost per mole.

Answer 1

"Cost" of "table salt" per "mole":

"Molecular mass" of "table salt" (NaCl) = 23 + 35.5 = 58.5

Now, "1000 g" of "NaCl cost" = Rs.2

Therefore 58.5 g of NaCl will cost =$\quad \frac { 1 }{ 1000 } \quad \times \quad 58.5\quad =\quad 0.117\quad rupees\quad =\quad 12paise\quad (approx)$

Cost of sugar per mole:

$Molecular mass of sugar ({ C }_{ 12 }{ H }_{ 22 }{ O }_{ 11 }) =\quad 12\quad \times \quad 12\quad +\quad 22\quad \times \quad 1\quad +\quad 11\quad \times \quad 16\quad =\quad 342$

Now, 100 g of sugar cost = Rs.6

$=\quad \frac { 6 }{ 1000 } \quad \times \quad 342\quad =\quad 2.052\quad rupees$

$=\quad 2.0\quad rupees (approx)$"

Answer 2

"Cost" of "table salt" per "mole":

"Molecular mass" of "table salt" (NaCl) = 23 + 35.5 = 58.5

Now, "1000 g" of "NaCl cost" = Rs.2

Therefore 58.5 g of NaCl will cost =\quad \frac { 1 }{ 1000 } \quad \times \quad 58.5\quad =\quad 0.117\quad rupees\quad =\quad 12paise\quad (approx)

1000

1

×58.5=0.117rupees=12paise(approx)

Cost of sugar per mole:

Molecular mass of sugar ({ C }_{ 12 }{ H }_{ 22 }{ O }_{ 11 }) =\quad 12\quad \times \quad 12\quad +\quad 22\quad \times \quad 1\quad +\quad 11\quad \times \quad 16\quad =\quad 342Molecularmassofsugar(C

12

H

22

O

11

)=12×12+22×1+11×16=342

Now, 100 g of sugar cost = Rs.6

=\quad \frac { 6 }{ 1000 } \quad \times \quad 342\quad =\quad 2.052\quad rupees=

1000

6

×342=2.052rupees

=\quad 2.0\quad rupees (approx)=2.0rupees(approx) "