Question

The molecular weight of glucose, \text{C}_6 \text H_{12} \text O_6C 6​ H 12​ O 6​ C, start subscript, 6, end subscript, H, start subscript, 12, end subscript, O, start subscript, 6, end subscript, is 180.16\,\dfrac{\text g}{\text{mol}}180.16 molg​ 180, point, 16, space, start fraction, g, divided by, m, o, l, end fraction.How many moles of glucose are in 19.1\,\text g19.1g19, point, 1, space, g of glucose?Express the answer using 3 significant figures. \text{mol}mol

Answer 1

• There are 0.106 mol of glucose.

Step-by-step explanation:

1) Data:

a) Chemical formula: C₆H₁₂O₆

b) Molecular weight, MM = 180.16 g/mol

c) mass, m = 19.1 g

c) n = ?

2) Formula:

• n = massi grams / MM

3) Soltuion:

Substitute the data:

• n = 19.1 g / 180.16 g/mol = 0.106 mol

The significant figures start with the first non-zero digit after the decimal point, so they are 1, 0, and 6, i.e. three significant figures.