A mixture of baking soda is prepared in water. A small amount of gas is released, weighing 5g. The mass of the remaining liquid mixture is 220g. The original mass of baking soda was 50g. What is the original mass of water to which the baking soda was added? Responses A 50g50g B 175g175g C 225g225g D 275g

1 answer

To find the original mass of water, we can set up the equation based on the conservation of mass.

Let:

  • \( m_{baking , soda} = 50 , g \) (mass of baking soda)
  • \( m_{gas} = 5 , g \) (mass of gas released)
  • \( m_{remaining , mixture} = 220 , g \) (mass of remaining liquid after gas is released)

We can use the following relationship:

\[ m_{original , mixture} = m_{baking , soda} + m_{water} \]

The original mass of the mixture was the mass of the baking soda plus the mass of the water. After the reaction, the mass of the remaining mixture is equal to the initial mass of the mixture minus the mass of the gas released:

\[ m_{remaining , mixture} = m_{original , mixture} - m_{gas} \]

By substituting the known values, we have:

\[ 220 , g = (50 , g + m_{water}) - 5 , g \]

Simplifying this equation gives us:

\[ 220 , g = 50 , g + m_{water} - 5 , g \] \[ 220 , g = 45 , g + m_{water} \]

Next, solve for the mass of the water:

\[ m_{water} = 220 , g - 45 , g \] \[ m_{water} = 175 , g \]

Thus, the original mass of water is 175 g. Therefore, the answer is:

B 175g.