Question

6 N HCL preparation

Answer 1

Let mass of one kind of tea x kg be mixed with another kind of tea mass y kg

SP of x kg tea = Rs 80x

Loss = (CP - SP)/CP

0•2 CP = CP - 80x

0•8CP = 80x. Or CP = 100x

Similarly

Profit = (SP - CP)/CP

0•25CP = 200y - CP

1•25CP = 200y . Or CP = 20000y/125 =160y

Total CP of x and y kg of tea = 100x + 160y

SP of x and y kg of tea = (x + y )×150

Profit = (SP - CP)/CP

0•25 = {(x + y )150 - (100x + 160y)}/(100x +160y)

0•25(100x + 160y) = (50x - 10y)

25x + 40y = 50x - 10y

25x = 50y

x/y = 2/1

Answer 2

Answer:

Explanation:

To prepare a 2L 6N HCl solution, you mix 1L of conc. HCl with 1L of milliQ water.