我也做到这道题了呵呵.
设原溶液中W有 x g ,水有 y g ,s 溶解度.
则由题意有:(x+6)/(y-10) = (x+1)/(y-30) = s/100
根据比例性质(关键):(x+6)/(y-10) = (x+1)/(y-30)
= [(x+6)-(x+1)]/[(y-10)-(y-30)] = 5/20 = 25/100
∴s=25 g/100g水.
我也做到这道题了呵呵.
设原溶液中W有 x g ,水有 y g ,s 溶解度.
则由题意有:(x+6)/(y-10) = (x+1)/(y-30) = s/100
根据比例性质(关键):(x+6)/(y-10) = (x+1)/(y-30)
= [(x+6)-(x+1)]/[(y-10)-(y-30)] = 5/20 = 25/100
∴s=25 g/100g水.