A laboratory stockroom has a stock solution of 90%(m/v) sulfuric acid,H2SO4(aq). If a technician dilutes 50ml of the stock solution to a final volume of 300ml, what will be the new mass/volume percent concentration?(HINT: the dilution formula can be used for concentration expressed in any units, provided that the units remain the same.)
Please help me i have no idea how to do this one:( i appreciate your time:) thankyou😺
Sure! I'm here to help. To solve this problem, we can use the dilution formula.
The dilution formula is expressed as follows:
C1V1 = C2V2
Where:
C1 = initial concentration
V1 = initial volume
C2 = final concentration
V2 = final volume
In this case, the initial concentration (C1) is 90%(m/v) or 0.9, the initial volume (V1) is 50mL, and the final volume (V2) is 300mL. We need to find the final concentration (C2).
Rearranging the formula to solve for C2, we get:
C2 = (C1 * V1) / V2
Plugging in the values, we have:
C2 = (0.9 * 50mL) / 300mL
C2 = 0.15
Therefore, the new mass/volume percent concentration of the diluted solution is 0.15 or 15%.