✔ 最佳答案
首先,解條題目:
34g的zine suplphate 於75cm^3的水內,那此溶液的濃度是多少?
in 'M' = 濃度單位 moles per cubic decimeter = mol dm ^-3 = Molar 統稱
到高程度時,濃度= M, uM, nM etc...M=1000uM 等.
但你這裡唔駛用.
首先知道zine suplphate = ZnSO4,再計它們1mole係幾多,以下:
Zn=65.39, S = 32.07, O4= 16.0x 4 = 64.0 在元素表mass找到.
全部加起= 65.39+32.07+64 = 161.46 (正常四捨五入,不過唔知你係唔係)
公式:
Mass in grams / mass of one mole (in grams)
*Mass in grams = 34g 如題
*mass of one mole (in grams) = 161.46
34/161.46 = 0.21mol
然後計在75cm^3水內有幾多濃度.
So in 1000cm^3, the number of moles = 0.21 x 1000/75
= 2.79993 mol
Concentration of ZnSO4 solution in 75 cm^3 of water = 2.79993 mol dm -3 or M.
其實做多啲同温書,好易!
Sorry,唔慣用中文解,請見諒!