Q1: A solution of specific gravity 1 consists of 35% A by weight and the remaining B. If the specific gravity of A is 0.7, the specific gravity of Bis

ANS:D - 1.2

To find the specific gravity of component B, we need to use the concept of specific gravity and weight percentage. The specific gravity of a substance is defined as the ratio of the density of that substance to the density of a reference substance (usually water). Since the specific gravity of component A is given as 0.7, it means that the density of A is 0.7 times that of water. Now, let's assume the density of water as 𝜌𝑤𝑎𝑡𝑒𝑟ρwater​. Then, the density of A (𝜌𝐴ρA​) can be calculated as: 𝜌𝐴=0.7×𝜌𝑤𝑎𝑡𝑒𝑟ρA​=0.7×ρwater​ Given that the solution has a specific gravity of 1, it means its density is the same as that of water. Therefore, the weight percentage of component B will determine its density: Weight percentage of B=100%−35%=65%Weight percentage of B=100%−35%=65% Now, since the specific gravity of the solution is 1, and component A constitutes 35% by weight, the density of component A should contribute to 35% of the overall density, and the density of component B should contribute to the remaining 65% of the overall density. Therefore, we can write: 0.35×𝜌𝐴+0.65×𝜌𝐵=𝜌𝑤𝑎𝑡𝑒𝑟0.35×ρA​+0.65×ρB​=ρwater​ 0.35×(0.7×𝜌𝑤𝑎𝑡𝑒𝑟)+0.65×𝜌𝐵=𝜌𝑤𝑎𝑡𝑒𝑟0.35×(0.7×ρwater​)+0.65×ρB​=ρwater​ 0.245𝜌𝑤𝑎𝑡𝑒𝑟+0.65×𝜌𝐵=𝜌𝑤𝑎𝑡𝑒𝑟0.245ρwater​+0.65×ρB​=ρwater​ 0.65×𝜌𝐵=𝜌𝑤𝑎𝑡𝑒𝑟−0.245𝜌𝑤𝑎𝑡𝑒𝑟0.65×ρB​=ρwater​−0.245ρwater​ 0.65×𝜌𝐵=0.755𝜌𝑤𝑎𝑡𝑒𝑟0.65×ρB​=0.755ρwater​ 𝜌𝐵=0.755𝜌𝑤𝑎𝑡𝑒𝑟0.65ρB​=0.650.755ρwater​​ Given that the density of water is 1 g/cm33, we can substitute this value: 𝜌𝐵=0.755×10.65ρB​=0.650.755×1​ 𝜌𝐵=0.7550.65ρB​=0.650.755​ 𝜌𝐵≈1.16 g/cm3ρB​≈1.16g/cm3 Therefore, the specific gravity of component B is approximately 1.161=1.1611.16​=1.16. So, none of the provided options seem to be correct. Let me know if you would like me to reconsider my approach or if you have any other questions!