Photo AI

The first four elements in group 1 are lithium, sodium, potassium and rubidium - Edexcel - GCSE Chemistry Combined Science - Question 5 - 2020 - Paper 1

Question icon

Question 5

The-first-four-elements-in-group-1-are-lithium,-sodium,-potassium-and-rubidium-Edexcel-GCSE Chemistry Combined Science-Question 5-2020-Paper 1.png

The first four elements in group 1 are lithium, sodium, potassium and rubidium. (a) Rubidium reacts with water to form rubidium hydroxide and hydrogen. $$ ext{2Rb(... show full transcript

Worked Solution & Example Answer:The first four elements in group 1 are lithium, sodium, potassium and rubidium - Edexcel - GCSE Chemistry Combined Science - Question 5 - 2020 - Paper 1

Step 1

Predict what you would see when a small piece of rubidium is placed in a large volume of water.

96%

114 rated

Answer

When a small piece of rubidium is placed in a large volume of water, you would observe vigorous fizzing or bubbling due to the rapid production of hydrogen gas. Additionally, there may be a significant exothermic reaction, which could result in flames or sparks as the rubidium reacts with water, leading to the formation of rubidium hydroxide.

Step 2

Why is rubidium more reactive than potassium?

99%

104 rated

Answer

The correct answer is D: rubidium has a more exothermic reaction with water than potassium does. This is indicative of its greater reactivity, as it indicates that the reaction releases more energy compared to potassium.

Step 3

Calculate the concentration of the rubidium hydroxide solution in g dm³.

96%

101 rated

Answer

To find the concentration, we first need to determine the number of moles of rubidium that reacted:

  1. Calculate the number of moles of rubidium:

    extmoles=massmolar mass=8.5g85g/mol=0.1mol ext{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{8.5 g}{85 g/mol} = 0.1 mol

  2. Since the reaction produces 2 moles of RbOH for every mole of Rb, the moles of RbOH produced will also be 0.1 mol.

  3. Now, calculate the concentration using:

    concentration (g/dm3)=mass of solute (g)volume of solution (dm3)\text{concentration (g/dm}^3\text{)} = \frac{\text{mass of solute (g)}}{\text{volume of solution (dm}^3\text{)}}

  4. The mass of RbOH produced is:

    mass=moles×molar mass=0.1mol×102g/mol=10.2g\text{mass} = \text{moles} \times \text{molar mass} = 0.1 mol \times 102 g/mol = 10.2 g

  5. Therefore, the concentration is:

    concentration=10.2g2.5dm3=4.08g/dm3\text{concentration} = \frac{10.2 g}{2.5 dm^3} = 4.08 g/dm^3

Join the GCSE students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;