# chemical calculation

What is the concentration of a solution ,in M, formed by adding 34g of zinc sulphate to 75cm^3 of water?

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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 &#39;M&#39; = 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!

• Firstly, you shall clarify the unit of concentration, M is molarity, then molarity = no. of moles / volume in dm3

No. of moles = 34 g / molar mass of Zn

volume in dm3 = 75 / 1000

Then concentration in M = (34/molar mass of Zn) / (75/1000)