I think the answer is a. and d.
hope this helps!!
Solubility data of a certain solute with a certain solvent is empirical. There are constant values for this at varying temperatures. For KCl in water at 25°C, the solubility is 35.7 g/100 mL of water. When you compare this with the solubility data of KCl with ethanol, this means that KCl is more soluble in water than in ethanol. This is true because KCl is an ionic salt which is very soluble in water.
substitute: <span><span>t<span>1/2</span></span>=<span><span>ln(2)</span>k</span>→k=<span><span>ln(2)</span><span>t<span>1/2</span></span></span></span>
Into the appropriate equation: <span>[A<span>]t</span>=[A<span>]0</span>∗<span>e<span>−kt</span></span></span>
<span>[A<span>]t</span>=[A<span>]0</span>∗<span>e<span>−<span><span>ln(2)</span><span>t<span>1/2</span></span></span>t</span></span></span>
<span>[A<span>]t</span>=(250.0 g)∗<span>e<span>−<span><span>ln(2)</span><span>3.823 days</span></span>(7.22 days)</span></span>=67.52 g</span>
