An antiseptic solution is made by mixing phenol with water in the ratio 2: 25. This solution is then mixed with further water in the ratio of 3: 4. How much water would be contained in 1L of the final solution? (Answer to the nearest mL.)
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Ratios Question please help (1 Viewer)
- Thread starter kpad5991
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You need to work backwards. Also, though you want the amount of water, it is easier to find the amount of phenol and then subtract from the total volume as water is added in each stage. So, start with the product, a 1 L solution that was prepared from a 3:4 mixture of the diluted phenol and water. The amount of the phenol mixture in the 1 L is thus
This diluted phenol was prepared from phenol and water in a 2:25 ratio. So, the original phenol used was
The entire final 1 L is phenol and water, and we now know how much phenol there was, thus
 &= 1 - V(\text{phenol in 1 L mixture}) \\ &= 1 - \cfrac{2}{63} \\ &= \cfrac{63-2}{63} \\ &= \cfrac{61}{63}\ \text{L} \\ &= \cfrac{61}{63} \times 1000\ \text{mL} \\ &= 968.25...\ \text{mL} \\ &\approx 968\ \text{mL} \qquad \text{(nearest mL)} \end{align*})
This diluted phenol was prepared from phenol and water in a 2:25 ratio. So, the original phenol used was
The entire final 1 L is phenol and water, and we now know how much phenol there was, thus