How much water should be mixed with 237 ml of ammonia, whose strength is 100%, in order to create a mixture that is diluted to a 75% strength.

Let the x ml of water be mixed with 237 ml of ammonia whose strength is 100%, in order to create a mixture of 75% strength.

The equation required to solve this will be:

237 / (237+x) = 75/100

solving for x we get:

237 / (237+x) = 3/4

237*4=3 (237+x)

948=711+3x

3x+711=948

3x=948-711

3x=237

x=237/3

x=79 ml

therefore the amount of water to be added will be 79 ml

Let the amount of water to be mixed = x ml

new mixture = 237+x

The new mixture is diluted to 75% strength i. e. 75/100

75/100 = 3/4

Therefore,

237 / (237+x) = 3/4

4 * 237 = 3 (237+x)

948=711+3x

3x=948-711

3x=237

x=237/3

x=79

The amount of water to be mixed with ammonia = 79ml